Suppose that $X$ and $Y$ are independent beta random variables (r.v.'s) with parameters $(a,b)$ and $(c,d)$, respectively. Let 
\begin{equation*}
	V:=\frac X{X+Y}. \tag{0}
\end{equation*}
The transformation $(x,v)\mapsto(x,x\frac{1-v}v)$ transforms $(X,V)$ to $(X,Y)$. 
The Jacobian determinant of this transformation is $-x/v^2$. Also,
\begin{equation*}
\Big(x,x\frac{1-v}v\Big)\in(0,1)\times(0,1)
\iff
	(0 < v<1\ \&\ 0 < x < s),
\end{equation*}
where 
\begin{equation*}
	s:=1\wedge\frac1r=\min\Big(1,\frac1r\Big),\quad r:=r_v:=\frac{1-v}v. 
\end{equation*}
The joint pdf of $(X,Y)$ is given by 
\begin{equation*}
	f_{X,Y}(x,y)=Cx^{a-1}(1-x)^{b-1}y^{c-1}(1-y)^{d-1}\,I\{0<x<1\ \&\ 0<y<1\}, 
\end{equation*}
where $I$ denotes the indicator and 
\begin{equation*}
	C:=\frac1{B(a,b)B(c,d)}. 
\end{equation*}
So, joint pdf of $(X,V)$ is given by 
\begin{equation*}
	f_{X,V}(x,v)=f_{X,Y}(x,xr)x/v^2
	=
	Cx^{a+c-1}(1-x)^{b-1}v^{-c-1}(1-rx)^{d-1}\times\text{ind}, 
\end{equation*}
where 
\begin{equation*}
	\text{ind}:=I\{0 < v<1\ \&\ 0 < x < s\}. 
\end{equation*}
So, the pdf of $V$ is given by 
\begin{equation*}
	f_V(v)=\int_{\mathbb R}f_{X,V}(x,v)\,dx
	=Cv^{-c-1}J(v)I\{0 < v<1\}, \tag{1}
\end{equation*}
where 
\begin{equation*}
	J(v):=\int_0^s x^{a+c-1}(1-x)^{b-1}(1-rx)^{d-1}\,dx. 
\end{equation*}
To evaluate $J(v)$, use formula 3.197.3 of *Table of Integrals, Series, and Products,
Seventh Edition* by Gradshteyn and Ryzhik: 
\begin{equation}
	\int_0^1 x^{\lambda-1}(1-x)^{\mu-1}(1-\beta x)^{-\nu}\,dx
	=B(\lambda,\mu)\,_2F_1(\nu,\lambda;\lambda+\mu;\beta) \tag{*}
\end{equation}
for $\lambda>0$, $\mu>0$, $|\beta|<1$, where $_2F_1$ is the hypergeometric function. 

If $1/2<v<1$, then $0<r<1$, $s=1$, and (*) immediately yields 
\begin{equation}
	J(v)=B(a+c,b)\,_2F_1(1-d,a+c;a+c+b;r). \tag{2}
\end{equation}

If $0<v<1/2$, then $r>1$, $s=1/r\in(0,1)$, and the substitution $t=rx$ together with (*) 
yields 
\begin{equation}
	J(v)=r^{-a-c}\int_0^1 t^{a+c-1}(1-t/r)^{b-1}(1-t)^{d-1}\,dx
	=r^{-a-c}B(a+c,d)\,_2F_1(1-b,a+c;a+c+d;1/r). \tag{3}
\end{equation}

Formulas (1), (2) (for $v\in(1/2,1)$), and (3) (for $v\in(0,1/2)$) give the pdf $f_V$ of $V$, defined by (0).